Problem: Given the equation: $ y = 4x^2 + 16x + 11$ Find the parabola's vertex.
Solution: When the equation is rewritten in vertex form like this, the vertex is the point $({h}, {k})$ $ y = A(x - {h})^2 + {k} $ We can rewrite the equation in vertex form by completing the square. First, move the constant term to the left side of the equation: $ \begin{eqnarray} y &=& 4x^2 + 16x + 11 \\ \\ y - 11 &=& 4x^2 + 16x \end{eqnarray} $ Next, we can factor out a $4$ from the right side: $ y - 11 = 4(x^2 + 4x) $ We can complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $4$ , so half of it would be $2$ , and squaring that gives us ${4}$ . Because we're adding the $4$ inside the parentheses on the right where it's being multiplied by $4$ , we need to add ${16}$ to the left side to make sure we're adding the same thing to both sides. $ \begin{eqnarray} y - 11 &=& 4(x^2 + 4x) \\ \\ y - 11 + {16} &=& 4(x^2 + 4x + {4}) \\ \\ y + 5 &=& 4(x^2 + 4x + 4) \end{eqnarray} $ Now we can rewrite the expression in parentheses as a squared term: $ y + 5 = 4(x + 2)^2 $ Move the constant term to the right side of the equation. Now the equation is in vertex form: $ y = 4(x + 2)^2 - 5 $ Now that the equation is written in vertex form, the vertex is the point $({h}, {k})$ $ y = A(x - {h})^2 + {k} $ $ y = 4(x - {(-2)})^2 + {(-5)} $ The vertex is $({-2}, {-5})$. Be sure to pay attention to the signs when interpreting an equation in vertex form.